Now let P(R) be the Grothendieck group of all finitely-generated projective R-modules. Thus, P(R) is the free abelian group generated by [P] for finitely-generated P, modulo [P] = [Q] + [Q’] for each short exact 0 → Q’ → P → Q → 0. By the lemma here, this means Q is a direct summand of P so P ≅ Q⊕Q’.

Theorem. The group P(R) is the free abelian group generated by [P] for indecomposable finitely-generated projective modules P. Furthermore, if in P(R) for projective modules then

Proof

By Krull-Schmidt’s theorem, every finitely-generated module P is uniquely written as a direct sum of indecomposable modules; if P is projective, so is each direct summand. Hence [P] is a sum of [Q] for indecomposable projective Q. Furthermore, each short exact 0 → Q’ → P → Q → 0 gives a decomposition P ≅ Q⊕Q’ so [P] = [Q] + [Q’] if and only if the modules on the LHS and RHS match after decomposition. The general case of r>2 follows by induction on r. ♦

Now we define a map:

which takes a projective module P to its class [P] in K(R). Note that this is a well-defined group homomorphism. The next map we define is:

which is given by , where J := J(R) is the Jacobian radical of R. Note that this is well-defined since a short exact sequence 0 → Q’ → P → Q → 0 of projective modules splits to give P ≅ Q⊕Q’ and so P/JP ≅ (Q/JQ) ⊕ (Q’/JQ’), and we get [P/JP] = [Q/JQ] + [Q’/JQ’] in K(R). Furthermore, we saw earlier that gives a bijection between projective indecomposable modules and simple modules. Since P(R) is freely generated by the projective indecomposable modules while K(R) is freely generated by the simple modules, we have:

Theorem. The map is a group isomorphism.

The map can be represented by a matrix with integer entries; let’s compute this for a simple example.

Example

is a direct sum of indecomposable projective modules. On the other hand, K is isomorphic to the column of 3-vectors, which has a composition series: Denoting the consecutive factors by A, B, C, we see that in K(R), we have [I] = [A], [J] = [A]+[B], [K] = [A]+[B]+[C], so the matrix is:

Exercise

Calculate the corresponding matrix for the ring:

Pairing

Next, consider the pairing given by:

We claim that this is well-defined; indeed for the first term, an exact sequence of projective modules splits as P ≅ Q⊕Q’ so Hom(P, M) is the direct sum of Hom(Q, M) and Hom(Q’, M), and the RHS gives:

On the other hand, if 0 → M’ → M → M” → 0 is an exact sequence of modules, then since P is projective, so is the resulting: